Question

A student was tasked with creating 450.0 mL of a 1.75 M HCl solution from a 4.50 M HCl stock solution.

How much of the stock solution should be transferred in order to conduct the dilution?

A. 1,160 mL
B. 1.16 L
C. 175L
D. 0.175L

Answer 1

Option D is correct.

The volume of the 4.50 M stock solution required to be diluted to the specifications given is 0.175 L.

Step-by-step explanation:

The law of dilution gives that the number of moles in both solutions stay the same. Mathematically, the law is given as

C₁V₁ = C₂V₂

The student wants to make 450 mL of 1.75 M of HCl.

We can find the number of moles of HCl in this required solution.

Number of moles = (Concentration in mol/L) × (Volume in L)

Concentration in mol/L = 1.75 M

Volume in L = (450/1000) = 0.45 L

Number of moles = 1.75 × 0.45 = 0.7875 mole

The stock solution has a concentration of 4.50 M

Volume of Stock solution needed

= (Number of moles)/(Concentration of stock solution in mol/L)

= (0.7875/4.5) = 0.175 L

Hence, the volume of the 4.50 M stock solution required to be diluted to the specifications given is 0.175 L.

Hope this Helps!!!